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Minimal Special Relativity - Reciprocity of Time Dilation

  1. Oct 25, 2014 #1


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    I hope somebody can help me with this question.

    1. The problem statement, all variables and given/known data
    I have to prove that time dilation is a reciprocal effect by only using the Postulates of Minimal Relativity Theory:
    1. Let I be an inertial frame, then I' moving with a constant velocity in a rectilinear motion with respect to I will also be an inertial frame.
    2. The transformation from I to I' is continuous in v.
    3. The motion is reciprocal. If I' moves with v wrt to I then I moves with -v wrt to I'

    2. Relevant equations

    I formulated transformations from I to I' through the following equations:
    Where x,x',v,v' are vector quantities.
    Now here are a few questions I have regarding some assumptions we can make about the transformations T and X:
    I read that one of the conditions for an inertial frame is that it describes time and space homogeneously and isotropically. My question is whether the transformation T and X will then have to be linear mappings in x and t to conserve homogeneity and isotropy? Is there non-linear mapping which fulfils these criterions?

    3. The attempt at a solution
    So basically i used the relativity principle and differentiated the x and x' equation wrt to t and t' respectively.
    Where x is the position of frame I wrt. I' and x' the position of frame I' wrt I.

    [itex]\frac{dx'}{dt'}=\left(v\nabla X(x,v,t)+\frac{\partial X(x,v,t)}{\partial t}\right)\frac{dt}{dt'}[/itex]
    [itex]\frac{dx}{dt}=\left(-v\nabla ' X(x',-v,t')+\frac{\partial X(x',-v,t')}{\partial t'}\right)\frac{dt'}{dt}[/itex]
    By the principle of relativity we know that:
    And I bascially want to prove:
    Unity might be a solution but there is another solution which seems paradoxical, but is actually not if you keep in mind that t and t' each mean two different things on both sides of the equation.

    If I don't rotate I' wrt to I then [itex]\nabla X(x,v,t)=\nabla ' X(x',-v,t')[/itex]
    My guess is that if i rotated it then I would also have to rotate v in the other direction and the rotation matrices would cancel. Then somehow I need to show that:
    [itex]\frac{\partial X(x,v,t)}{\partial t}=-\frac{\partial X(x',-v,t')}{\partial t'}[/itex]
    From the Lorentz transformation I know that this is true. since the only term which contains a t in the x' transformation is the vt term. But I don't know whether it is a valid argument to say that X will have to be a linear combination of terms of which one is vt.
  2. jcsd
  3. Oct 30, 2014 #2


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    Staff Emeritus
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    The two relevant equations here are:
    1. t' = T(x,v,t)
    2. t = T(x', -v, t')
    So if we use 1. to compute [itex]\frac{dt'}{dt}[/itex], then we get:
    [itex]\frac{dt'}{dt} = \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial v} \frac{d v}{d t} + \frac{\partial T}{\partial t}[/itex]
    [itex] = \frac{\partial T}{\partial x} \frac{dx}{dt} + \frac{\partial T}{\partial t}[/itex] (because [itex]v[/itex] is constant)

    If we use 2. to compute [itex]\frac{dt}{dt'}[/itex], then we get something similar.

    The only things left to figure out are two quantities: [itex]\frac{dx}{dt}[/itex] and [itex]\frac{dx'}{dt'}[/itex]. But you have to think about what this means. In the first case, [itex]x(t)[/itex] is the x-position of a clock that is at rest in primed coordinate system. If it is at rest in the primed coordinate system, then what is its speed in the unprimed coordinate system? So what is [itex]\frac{dx}{dt}[/itex] for such a clock?

    In the second case, [itex]x'(t')[/itex] is the x'-position of a clock that is at rest in the unprimed coordinate system. Then what is [itex]\frac{dx'}{dt'}[/itex]?
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